No specific type is being verified by Type

Question

No specific type is being verified by Type

Below is a snippet of sample code:

interface Foo {
    bar: string;
    buz: string;
}
const fooList: Foo[] = [
    { bar: '1', buz: '1' },
    { bar: '2', buz: '2' }
];
const newList: Foo[] = fooList.map((foo: Foo) => ({
    bar: foo.bar,
}));

Upon execution, an error message

Property 'buz' is missing in type '{ bar: string; }'

will be displayed.

If I modify the type of fooList to any, TypeScript does not show any errors:

interface Foo {
    bar: string;
    buz: string;
}
const fooList: any = [
    { bar: '1', buz: '1' },
    { bar: '2', buz: '2' }
];
const newList: Foo[] = fooList.map((foo: Foo) => ({
    bar: foo.bar,
}));

In this scenario, I anticipated receiving the same error as before. Why does TypeScript not raise an error in the second case?

typescript

Answer 1

Answer №1

The issue arises because the variable `fooList` is declared as type `any`, leaving the compiler unable to make any inferences regarding the `fooList.map` function (since it no longer recognizes `fooList` as an array).

As a result, the compiler cannot determine the return type of `map`, or how it handles the callback provided to `map`. Without this information, it does not recognize that the callback's output should be returned by `map` as an element within an array of type `Foo`, leading to a failure to alert you if the callback's output is not a valid `Foo` object.

To resolve this, consider changing the type of `fooList` to `any[]` instead of just `any`. This will inform the compiler that `fooList` is indeed an array, allowing it to understand the functionality of `fooList.map` and provide the appropriate error messages.

Alternatively, explicitly specify that your callback must return a `Foo` object:

const newList: Foo[] = fooList.map((foo: Foo): Foo => ({
    bar: foo.bar,
}));

This approach will enable the compiler to detect errors while still maintaining `fooList` as type `any`.

Answer 2

The issue arises because the variable `fooList` is declared as type `any`, leaving the compiler unable to make any inferences regarding the `fooList.map` function (since it no longer recognizes `fooList` as an array).

As a result, the compiler cannot determine the return type of `map`, or how it handles the callback provided to `map`. Without this information, it does not recognize that the callback's output should be returned by `map` as an element within an array of type `Foo`, leading to a failure to alert you if the callback's output is not a valid `Foo` object.

To resolve this, consider changing the type of `fooList` to `any[]` instead of just `any`. This will inform the compiler that `fooList` is indeed an array, allowing it to understand the functionality of `fooList.map` and provide the appropriate error messages.

Alternatively, explicitly specify that your callback must return a `Foo` object:

const newList: Foo[] = fooList.map((foo: Foo): Foo => ({
    bar: foo.bar,
}));

This approach will enable the compiler to detect errors while still maintaining `fooList` as type `any`.

Answer 3

Answer №2

Because any is truly unique. According to the official documentation, it serves as a means to

bypass type-checking and allow values to pass through compile-time checks.

By using any, you have the freedom to assign any value, access any property, call any method, without facing any objections from the compiler. It automatically infers the any type in most scenarios when dealing with operations involving it.

In the second example, if you don't provide a type annotation for newList, its type is automatically designated as any, making it compatible for assignment to Foo[].

Answer 4

Because any is truly unique. According to the official documentation, it serves as a means to

bypass type-checking and allow values to pass through compile-time checks.

By using any, you have the freedom to assign any value, access any property, call any method, without facing any objections from the compiler. It automatically infers the any type in most scenarios when dealing with operations involving it.

In the second example, if you don't provide a type annotation for newList, its type is automatically designated as any, making it compatible for assignment to Foo[].

No specific type is being verified by Type

Answer №1

Answer №2

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